\begin{answer}{normalfourthmoment}
Begin by integrating.
Derive the fourth moment,
\begin{align}
\E(X^4)
&= \int_{-\infty}^{\infty}{
          x^4
          f(x)
          dx} \nonumber \\
&= \int_{-\infty}^{\infty}{
          x^4
          \frac{1}{\sqrt{2 \pi} \sigma }
          \exp{\left( -\frac{x^2}{2\sigma^2}  \right)}
          dx}
\label{eq:normalfourthmoment:integraltosolve}
\text{.}
\end{align}
There doesn't appear to be a quick way to do this integral, so use integration by parts:
\[
\int_{-\infty}^{\infty}{ u dv }
=
uv\Big\vert_{x=-\infty}^{x=\infty} - \int_{-\infty}^{\infty}{ v du }
\text{.}
\]
The best way to rewrite
(\ref{eq:normalfourthmoment:integraltosolve}) in order to choose $u$ and $v$ is by trial and error.
Be sure to tell the interviewer what you are doing while you are writing out the formula for integration by parts, and mention that you are trying to find the best candidates for $u$ and $v$.
The only candidate that gives you something \emph{easier} to deal with after integration by parts is
\begin{align*}
\E(X^4)
&=
          \frac{1}{\sqrt{2 \pi} \sigma }
          \int_{-\infty}^{\infty}{
          x^3
          \;
          x
          \exp{\left( -\frac{x^2}{2\sigma^2}  \right)}
          dx}
\text{.}
\end{align*}
Then you have
\begin{align*}
u   &= x^3      \\
dv  &=
        x
        \exp{\left( -\frac{x^2 }{2\sigma^2} \right)}
        dx
        \\
%
du  &= 3 x^2 dx \\
v   &=
       - \sigma^2
        \exp{\left( -\frac{x^2}{2\sigma^2}  \right)}
\end{align*}
giving
\begin{equation}
\begin{aligned}
\label{eq:fourthmoment:nearlythere}
\frac{1}{\sqrt{2 \pi} \sigma }
\int\limits_{-\infty}^{\infty}{
  x^3
  \;
  x \exp{\left( -\frac{x^2}{2\sigma^2}  \right)}
}
=
&-\frac{1}{\sqrt{2 \pi} \sigma }
x^3
\sigma^2
\exp{\left( -\frac{x^2}{2\sigma^2}  \right)} \bigg|_{x=-\infty}^{x=\infty}
\\
&+
\frac{1}{\sqrt{2 \pi} \sigma }
\int\limits_{-\infty}^{\infty}{
  \sigma^2
  \exp{\left( -\frac{x^2}{2\sigma^2}   \right)}
 3 x^2
  dx
}
\text{.}
\end{aligned}
\end{equation}
The first term on the right hand side is equal to zero.
Use L'Hôpital's rule to prove it.%
\footnote{But be wary:
I once got reprimanded during an interview at a hipster hedge fund for suggesting
L'Hôpital's rule as a solution to a problem that had the form $\nicefrac{\infty}{\infty}$.
The interviewer had never heard of L'Hôpital, and snidely enquired whether it was some sort of food.
I was impressed that someone managed to graduate with a physics degree without ever encountering L'Hôpital.
The interviewer must have been great at brainteasers to get hired.}
Apply the rule three times:
\begin{equation}
\label{eq:fourthmoment:lhopital}
\begin{aligned}
\lim_{x \rightarrow \infty}
-x^3
\sigma^2
\exp{\left( -\frac{1}{2\sigma^2} x^2 \right)}
&=
\lim_{x \rightarrow \infty}
\frac{-x^3 \sigma^2 }{ \exp{\left( \frac{1}{2\sigma^2} x^2 \right)} }
\quad\text{rewrite as fraction}
\\
&=
\lim_{x \rightarrow \infty}
\frac{
  - 3 \sigma^{2} x^{2}
  }{
  \quad \frac{x}{\sigma^{2}} \exp{\left(\frac{x^{2}}{2 \sigma^{2}}\right)}
  }
\quad\text{L'Hôpital 1}
\\
&=
\lim_{x \rightarrow \infty}
\frac{ - 6 \sigma^{2} x }{
\frac{1 }{\sigma^{2}} \left(1 + \frac{x^{2}}{\sigma^{2}}\right)
\exp{\left(\frac{x^{2}}{2 \sigma^{2}}\right)}
}
\quad\text{L'Hôpital 2}
\\
&=
\lim_{x \rightarrow \infty}
\frac{ - 6 \sigma^{2} }{
\frac{x}{\sigma^{4}} \left(3 + \frac{x^{2}}{\sigma^{2}}\right)
\exp{\left(\frac{x^{2}}{2 \sigma^{2}}\right)}
}
\quad\text{L'Hôpital 3}
\\
&= 0
\end{aligned}
\end{equation}
and the same is true for
${x \rightarrow -\infty}$.
While I include the elaborate calculations above, you don't have to, because it is clear that the denominator will always include a term involving
\[
\exp{\left(\frac{x^{2}}{2 \sigma^{2}}\right)}
\]
while the nominator will eventually become zero after enough differentiation.
So now (\ref{eq:fourthmoment:nearlythere}) becomes
\begin{align*}
\frac{1}{\sqrt{2 \pi} \sigma }
\int\limits_{-\infty}^{\infty}{
  x^4
  \exp{\left( -\frac{x^2}{2\sigma^2}  \right)}
  dx
}
&=
3 \sigma^2
\frac{1}{\sqrt{2 \pi} \sigma }
\int\limits_{-\infty}^{\infty}{
  x^2
  \exp{\left( -\frac{x^2}{2\sigma^2}  \right)}
  dx
}
\text{.}
\end{align*}
The right hand side is equal to $3\sigma^2$ multiplied by the second moment of a normal distribution with mean zero and standard deviation $\sigma$.
Recall that if $X \sim \text{Normal}(\mu, \sigma^2)$ you have
\begin{align*}
  \Var(X) &= \E(X^2) - \E(X)^2 \\
\sigma^2  &= \E(X^2) - \mu^2
\end{align*}
meaning
$ \E(X^2) = \sigma^2 $
when
$\mu=0$.
Thus the fourth moment of
$X \sim \text{Normal}\left(0, \sigma^2\right)$ is
\begin{align*}
\E(X^4)
&=
3 \sigma^2 \E(X^2)
\\&=
3 \sigma^4
\text{.}
\end{align*}
This was a very long journey to arrive at the answer, and doing the integration may not have been the best approach.
This approach does, however, allow you to exhibit your integration prowess, which may hurt or help you depending on how the interviewer feels.
In my case, I wasn't able to arrive at the right answer, but when I reached  (\ref{eq:fourthmoment:nearlythere}) the interviewer said ``there, now we have something familiar'' (referring to the second moment) and went on to the next question.
A better approach might have been to use the Moment Generating Function of the normal distribution,
\begin{align*}
  M(t) =
  e^{\mu t}
  e^{\frac{1}{2}\sigma^2 t^2}
  \text{.}
\end{align*}
Then you can calculate any moment using
\[
E \left( X^n \right) = M_X^{(n)}(0) = \left. \frac{d^n M_X (t)}{dt^n}\right|_{t=0}
\]
where $M_X^{(n)}(t)$ is the $n$th derivative of the function $M(t)$.
In the current case where $\mu=0$,
\begin{align*}
  M_X(t) &=
  e^{\frac{1}{2}\sigma^2 t^2}\\
M_X^{(1)}(t) &=
\sigma^{2} t e^{\frac{\sigma^{2} t^{2}}{2}}
\\
M_X^{(2)}(t) &=
\sigma^{2} \left(\sigma^{2} t^{2} + 1\right) e^{\frac{\sigma^{2} t^{2}}{2}}
\\
M_X^{(3)}(t) &=
\sigma^{4} t \left(\sigma^{2} t^{2} + 3\right) e^{\frac{\sigma^{2} t^{2}}{2}}
\\
M_X^{(4)}(t) &=
\sigma^{4} \left(\sigma^{4} t^{4} + 6 \sigma^{2} t^{2} + 3\right) e^{\frac{\sigma^{2} t^{2}}{2}}
\\
M_X^{(4)}(0) &=
3\sigma^{4}
\text{.}
\end{align*}
Still tedious, but perhaps less so than the integration.%
\footnote{I used Sympy by \citet{Sympy} to generate these, and for all the differentiation in (\ref{eq:fourthmoment:lhopital}).}
Of course, you could just learn the moments of the normal distribution by heart, but I don't think that's what interviewers want---they want to see some mathematics.
In my case, the interviewer also asked me what the fourth moment meant.
I didn't understand, but they probably wanted me to mention the kurtosis, which is a measure of the ``tailedness'' of a distribution.\
The fourth moment is used to measure kurtosis,
the third moment is related to the skewness,
the second moment to the variance, and the first moment is the mean.

Robert Tillman suggested the following clever solution, which is the shortest method to derive $E\left( X^4 \right)$ for the given normal distribution.
Recognise
\begin{align*}
\Var(X^2) &= \E\left((X^2)^2\right) - \E\left( X^2 \right)^2  \\
          &= \E\left(X^4\right) - \E\left( X^2 \right)^2  \\
\E\left(X^4\right)
&=  \Var(X^2) + \E\left( X^2 \right)^2
\end{align*}
and that
\begin{align*}
\frac{X^2}{\sigma^2} \sim \chi^2_{1}
\text{.}
\end{align*}
The $\chi^2_k$ distribution has a mean of $k$ and variance of $2k$.
Define a random variable
$Z \sim \chi^2_{1}$
and write
\begin{align*}
X^2 &= {\sigma^2} Z \\
\E\left(X^2 \right) &= {\sigma^2} \E( Z ) \\
                    &= {\sigma^2} (1) \\
\Var\left(X^2 \right) &= (\sigma^2)^2 \Var( Z ) \\
                    &= {\sigma^4} (2) \\
\end{align*}
giving the same answer as above,
\begin{align*}
\E\left(X^4\right) &= 2 \sigma^4 + \sigma^4 \\
                   &= 3 \sigma^4
\text{.}
\end{align*}


In my experience, interviewers from investment banks love questions about the normal distribution.
This might be in response to interview solution manuals like
\citet{HeardOnTheStreet} and \citet{JoshiQA},
and the fact that they don't contain many questions on the theory of the normal distribution.
Perhaps there is a book called ``Interesting Proofs about the Normal Distribution'' doing the rounds at the banks, or maybe there are internal study groups who meet regularly to make and solve brainteasers about the normal distribution.
Whatever it is, I cannot guarantee that these questions will remain popular.
I nonetheless recommend that you know the normal distribution well---not only because it might help you during interviews, but also because it is, by now, an overused distribution that you are sure to encounter in practice.
\end{answer}
